1
RL Circuits
Equipment DataStudio, RLC circuit board, 2 voltage sensors (no alligator clips), 2 leads
(35 in)
Reading Review operation of oscilloscope, signal generator, and power amplifier II
1
Introduction
The three basic linear circuit elements are the resistor, the capacitor, and the inductor. This
lab is concerned with the characteristics of inductors and circuits consisting of a resistor and
an inductor in series (RL circuits). The primary focus will be on the response of an RL
circuit to a step voltage and a voltage square wave.
2
2.1
Theory
Inductors
An inductor is a 2-terminal circuit element that stores energy in its magnetic field. Inductors
are usually constructed by winding a coil with wire. To increase the magnetic field inductors
used for low frequencies often have the inside of the coil filled with magnetic material. (At
high frequencies such coils can be too lossy.) Inductors are the least perfect of the basic
circuit elements due to the resistance of the wire they are made from. Often this resistance
is not negligible, which will become apparent when the voltages and currents in an actual
circuit are measured.
If a current I is flowing through an inductor, the voltage VL across the inductor is proportional to the time rate of change of I, or dI
. We may write
dt
VL = L
dI
,
dt
(1)
where L is the inductance in henries (H). The inductance depends on the number of turns of
the coil, the configuration of the coil, and the material that fills the coil. A henry is a large
unit of inductance. More common units are the mH and the µH. A steady current through
a perfect inductor (no resistance) will not produce a voltage across the inductor. The sign
of the voltage across an inductor depends on the sign of dI/dt and not on the sign of the
current. Figure 1 shows the relationship between the current and voltage for a resistor and in
inductor. The arrow indicates the positive direction of the current through the two devices
(resistor and inductor). That is, if the current is in the direction of the arrow, then I > 0.
[Note that just because the arrows point down in the Fig. 1 doesn’t mean that the current is
flowing in the downward direction. The current can also flow in the upwards direction, but
in that case I < 0.] VR and VL are the potential differences across the resistor and inductor,
respectively, defined such that if V(R,L) > 0, then the potential at the “+00 end of the device
is greater than the potential at the “−00 end. [According to this convention, if V(R,L) < 0
then the potential at the bottom is greater than the potential at the top.]
According to these conventions, for the case of the resistor, V = IR. If the current I > 0
(current flowing downward) the potential is greater at the top than at the bottom in Fig. 1.
2
Figure 1: Relationship between current and voltage in a resistor and inductor.
In the case of the inductor, if dI/dt > 0 then the potential is greater at the top of the
inductor than at the bottom. In the inductor, one can have the situation where the current
is flowing downward (I > 0), but decreasing in magnitude so that dI/dt < 0. In this case
VL = dI/dt < 0, so the potential at the top is smaller than at the bottom. The potential
is always lower on the “downstream” end of a resistor, but for an inductor, it
can be larger or smaller, depending on the rate of change of the current.
As mentioned above, the coil of wire making up an inductor has some resistance RL . This
resistance acts like a resistor RL that is in series with an ideal inductor. (An ideal inductor
is one with no resistance.) In this situation, the voltage across the inductor is
VL = L
2.2
dI
+ RL I.
dt
(2)
RL Circuits
A series RL circuit with a voltage source V (t) connected across it is shown in Fig. 2. The
Figure 2: RL circuit: The loopy arrow indicates the positive direction of the current. The + and − signs
indicate the positive values of the potential differences across the components.
voltage across the resistor and inductor are designated by VR and VL , and the current around
the loop by I. The signs are chosen in the conventional way; I is positive if it is in the
direction of the arrow. Also, the positive signs of the voltages across the various components
are indicated by the + and − symbols. Kirchoff’s law, which says that sum of the voltage
3
changes around the loop is zero, may be written
VL + VR = V,
(3)
where it is assumed that the voltages V (t), VR (t), and VL (t) can all vary with time t.
Assuming RL = 0 (any resistance in the inductor can just be added to that of the resistor)
and letting VL = L dI/dt and VR = IR, Equation (3) becomes
L
dI
+ RI = V.
dt
(4)
t
The solution to the homogeneous equation [V (t) = 0] is I(t) = I0 e− L/R , where I0 is the
t
current through the circuit at time t = 0. This solution leads immediately to VR = I0 Re− L/R
t
and VL = −I0 Re− L/R . The homogeneous solution decays exponentially with a time constant
τ = L/R.
Of interest is the response of an RL circuit to an abrupt change in the voltage V of the
source from V = 0 to a constant voltage V = V0 . Before the change, we assume that the
t
voltage of the source was at V = 0 for a long time (much longer than τ ). The function e− L/R
describes the time dependence of the circuit. To find the behavior as a function of time, it
is only necessary to use physical intuition and put the appropriate two constants into the
solution. One guiding principle is that the current I, and therefore VR does not change the
instant the voltage across an RL circuit is changed. That is, VR and I are continuous. This
means initially, that the entire voltage change must appear across the inductor.
QUESTION: What is the basis for the argument that I doesn’t change discontinuously
when the source voltage is suddenly changed from V = 0 to V = V0 ?
A solution to the inhomogeneous equation
L
dI
+ RI = V0
dt
(5)
is I = V0 /R. The general solution in this case is a linear combination of the homogenous
and inhomogeneous solutions that satisfy the initial and final conditions. This is
I=
t V0 1 − e− L/R .
R
(6)
This says that after the constant voltage V0 is applied, the current in the circuit will exponentially approach V0 /R if the inductor has no resistance, or V0 /(R + RL ) if the inductor
has a resistance RL . These considerations apply whatever the previous history of the RL
circuit. Consider an RL circuit where V = 0 and there is no current. We assume that RL =
0. If at t = 0 a constant voltage V0 is put across the circuit, for t ≥ 0, VL and VR are given
by,
t
t VL = V0 e− L/R
and
VR = V0 1 − e− L/R .
(7)
This behavior is illustrated in Fig.3. The current is not plotted, but remember that the
current is proportional to VR . Initially all of V0 appears across L because the current just
before and after the application of V0 is zero. As the current exponentially builds up the
voltage across the resistor increases and the voltage across the inductor decreases. If we wait
4
V0
VR
Voltage VR across the resistor
0
V0
VL
Voltage VL across the inductor
0
0
time
Figure 3: Voltages VR across the resistor R and VL across the inductor L as a function of time after the
source voltage is switched from V = 0 to V = V0 .
many time constants τ = L/R, the voltages reach steady state, and we have VR ∼
= V0 and
∼
VL = 0. If now V is set equal to 0 (this is equivalent to shorting the circuit) VL and VR will
be given by
t
t
VL = −V e− L/R
and
VR = V e− L/R .
(8)
The response of an RL circuit to an alternating pair of constant voltages, first at V0 , and then
0, can be observed by applying a square wave to the circuit, alternating between V = V0 and
V = 0. Figure 3 shows half the period of such an oscillation for the case where the period T
is much larger than the time constant τ = L/R.
QUESTION: How would Fig. 3 be modified if the inductor had some resistance?
QUESTION: If T L/R, what is the response of an RL circuit to a symmetric square
wave that oscillates between +V and −V (assume RL = 0)?
If a high frequency square wave, such that T L/R, is applied to an RL circuit, the
changes in current and VR are minimal. There is not enough time for the current through the
inductor to change much before the voltage is reversed. If the square wave is not symmetric
with respect to ground the average VR will be the average voltage of the square wave,
assuming RL = 0. Fig. 6 shows the voltages for a square wave that oscillates between
the constant voltage V and 0 (ground). The exponential dependences of VL and VR are
approximated as straight lines and it has been assumed that RL = 0 and T L/R.
5
3
3.1
Procedure
Measuring RL
In this section the DC value of RL will be measured. In the right experiment setup window,
click analog plug channel C and choose power amplifier. Click the AUTO button in the
signal generator window that appears. Program the signal generator for 2 V DC. Drag the
digits display icon to Output Current under Data. This digits display will give the current
delivered by the power amplifier. Program the digits display for 2 decimal places. Connect
the output of the power amplifier directly across the coil (see Fig. 4) and click Start and
then STOP. From your data calculate RL , the resistance of the coil. Click the FILE menu
Inductor
Resistors
Inductor
Core
Figure 4: Picture of the circuit board
button, NEW ACTIVITY, and DON’T SAVE. Remove the wires from the coil.
During your experiments compare the value of RL to the values of the resistances used in
the RL circuits.
3.2
Time-Dependent Voltages
In this section the response of an RL circuit will be examined experimentally using the
DataStudio signal generator and power amplifier II, two voltage sensors, and the oscilloscope
display. Click analog plug channel C and choose power amplifier. Click the AUTO button
in the signal generator window that appears. Drag the scope display icon to Output Voltage
under Data. The oscilloscope display will open with the top or green trace channel having
the signal generator voltage (analog output) as the input. Click analog plug channel A and
choose voltage sensor. Do this again for channel B. Program the middle scope channel (red
trace) for channel A and the bottom scope channel (blue trace) for channel B.
In the experiment various resistors will be connected in series with an 8.2 mH inductor.
Use the circuit board shown in Fig. 4 to assemble the circuit in Fig. 2. Connect the voltage
source with the polarity shown in Fig. 2. The voltage sensors should also be connected to the
6
resistor and inductor with the polarities shown in Fig. 2. The polarities are very important
in this experiment. A positive voltage from a voltage sensor will correspond to a positive
voltage as given by the convention of Fig. 2. The input square wave to the RL circuit should
be in the top scope channel (green trace), VL should be on the middle scope channel (red
trace), and VR on the bottom scope channel (blue trace).
QUESTION: There are buttons on the far right of the scope display that add adjacent
channels. In the experiments try adding the middle (VL ) and bottom (VR ) scope channels.
Do they match the output signal voltage? Why?
You may wonder about the particular choice of frequencies used in the experiments. It
turns out that the results are better if the signal generator does not operate at too high a
frequency and the time constants and frequencies have been chosen accordingly.
Some remarks.
• When you click STOP, the last traces are stored. The stored traces are better for
examination than the “live” traces as they are steady.
• You can use the smart cursor on the stored traces.
• You can use many of the scope functions on the stored traces.
• You can print out the stored traces.
• Label your printed out traces immediately by V , VL , and VR as the printers are not
color printers.
• Use the same vertical sensitivity for all 3 traces so that you can easily compare the trace
values.
• The default voltage amplitude of 5 V for the signal generator is NOT satisfactory for
the experiments. Use 2 V. (The current rating for the inductor is 0.8 A.)
• For reliable triggering with a positive only square wave, make the trigger voltage positive
but less than 2 V.
• Due to RL , your graphs will differ somewhat from Figs. 2-4 of this write-up, which
assumes RL = 0.
3.2.1
T L/R, Positive Only Square Wave
Hook up an RL circuit with the R = 33 Ω resistor. Compute the expected value of the time
constant L/Rtot (with uncertainty) for two cases: RL = 0 and RL given by your result from
the previous section. How do these values compare with the period (T ) of a 300 Hz square
wave?
Examine VL , VR , and V using a 300 Hz square wave that oscillates between ground (0 V)
and 2V. Be sure to set the sample rate to 5000 Hz. Are the traces what you expect?
Now to determine the time constant from the data. We want to plot one “charging” cycle
in Python. In DataStudio create a table recording VL and V . Write down the sampling rate,
and RECORD data for a very short time (less than one second). In order to record data, it
is necessary to close the scope display, then click start and stop almost immediately. Click
7
File, Export Data and choose your desired run from the voltage sensor which is across the
conductor. Save the file. Repeat for the data of Output Voltage. Open the data file in a
text editor and look for one “chunk” of data with V ≈ 2V. Copy these lines into a new file,
this will be the data used for plotting.
t
We expect VL = V e− R/L . Using your data, make a plot of log(VL ) versus t in Python,
and determine the slope and y-intercept (with uncertainties). What do you expect the slope
to be? The y-intercept? From your slope and intercept, compute the time constant and V .
Do they agree with the expected values? Which expression for the time constant is best to
use: L/R or L/(R + RL )?
3.2.2
T L/R, Positive Only Square Wave, Metal Core
Use the same setup as the previous section. With the DataStudio scope running, insert the
metal rod into the inductor coil. How is the trace changed? What parameter is the metal
rod changing?
With the rod inside the coil, redo the analysis from the previous section. After closing
the scope display, record a table of VL and V for a short time, export the table, and copy
out one section of data with V ≈ 2V. Plot log(VL ) versus t (the sampling rate should be the
same), and determine the slope and intercept. Are the numbers the same as in the previous
section?
Use the slope to calculate the new inductance L0 ± δL0 , assuming the resistance Rtot is
the same as in the previous section. How significant is the effect of the metal core?
3.2.3
T L/R, Symmetric Square Wave
Use the same parameters as in section 3.2.1 except use a square wave that is symmetric with
respect to ground (varies from 2V to -2V).
The time constant τ = L/Rtot is defined to be the time it takes VL to reduce by a factor of
e (=2.718...). Knowing this, we can approximate the time constant visually with the Smart
Cursor. In DataStudio, click Start then click STOP to freeze the trace. Find the peak of
the VL curve, and record the voltage Vpeak and the time tpeak . If we assume RL = 0, then VL
should decay to 0 with time. Using the Smart Cursor, find the point on the VL curve where
VL = Vmax /e, remember 1/e ≈ 0.37. Record the time of this point t1 . Then τ = t1 − tpeak ,
simple!
Compare your results to what you expect. Due to limitations in the equipment used and
the non-zero value of RL , the peak voltages will be less than calculated.
3.2.4
T L/R, Positive Only Square Wave
Apply a 3,000 Hz positive only square wave to an RL circuit with R = 10 Ω. Change the
x-axis interval to .5 ms/div and set the sample rate to 50,000 Hz. What is the expected
time constant and how does it compare to T ? Compare your results to Fig. 3. In Fig. 3,
is approximating the exponentials by straight lines reasonable? Note the distortion in the
square wave.
8
3.2.5
T L/R, Symmetric Square Wave
Use the same parameters as in section 3.2.4 except use a square wave that is symmetric with
respect to ground. Compare to Fig. 4. The effect of RL is quite small here as the average
current is small.
3.2.6
T ≈ L/R, Symmetric Square Wave
Apply a 500 Hz square wave to an RL circuit with R = 10 Ω. Are the results what you
expect? If you have time, try somewhat lower and higher frequencies.
3.2.7
RLC Circuit (Optional)
Make a series circuit with the 33 Ω resistor, the inductor, and the 330 µF capacitor. This is
called an RLC circuit, for obvious reasons. Use the voltage sensors to monitor the voltage
across the inductor and the capacitor (we will ignore the resistor). Monitor the voltages
in the DataStudio oscilloscope, and apply a 300 Hz, 2V, positive-only square wave to the
circuit. Change the sample rate to 5000 Hz. Describe what you see.
4
Finishing Up
Please leave the lab bench as you found it. Thank you.
9
Figure 5: The signal in the first row is the voltage across the inductor. The signal in the second row is the
voltage across the resistor and the bottom row is the voltage supply.
Figure 6: The signal in the first row is the voltage across the inductor. The signal in the second row is the
voltage across the resistor and the bottom row is the voltage supply.